Question

Show that the average field inside a sphere of radius R, due to all the charge within the sphere, is

${\mathit{E}}_{\mathbf{a}\mathbf{v}\mathbf{e}}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{\rho }}{{\mathbf{R}}^{\mathbf{3}}}$

Where $\mathit{\rho }$is the total dipole moment. There are several ways to prove this delightfully simple result. Here's one method:

(a) Show that the average field due to a single chargeqat point r inside thesphere is the same as the field at r due to a uniformly charged sphere with

$\mathit{\rho }\mathbf{=}\mathit{q}\mathbf{/}\left(\frac{4}{3}\pi R^3\right)$, namely

$\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}\left({\displaystyle \frac{4}{3}}\pi R^3\right)}\mathbf{\int }\frac{\mathbf{q}}{{\mathbf{r}}^{\mathbf{2}}}\mathit{r}\mathit{d}\mathit{\zeta }\mathbf{\text{'}}$

Where r is the vector from r to $\mathit{d}\mathit{\zeta }$

(b) The latter can be found from Gauss's law (see Prob. 2.12). Express the answerin terms of the dipole moment of q.

(c) Use the superposition principle to generalize to an arbitrary charge distribution.

(d) While you're at it, show that the average field over the volume of a sphere, dueto all the charges outside, is the same as the field they produce at the center.

Short answer

Answer

1. $E_{ave}=E_{pp}$is proved.

2. The expression for the electric filed due to dipole moment $E_{\rho }=-\frac{\rho }{4\pi {\epsilon }_0{R}^3}$.

3. The individual average electric filed is localid="1655725443659" $E_{\rho }=\frac{\rho }{4\pi {\epsilon }_0{R}^3}$.

4. The expression for filed at P due to uniformly charged sphere.

$E_{\rho }=-\frac{1}{4\pi {\epsilon }_0}\frac{-q}{r^2}r$

Step-by-step solution

Define functions

Consider the following figure,

The figure shows the sphere, r is the radius of sphere, the point charge inside the sphere.

Determine (a)

a)

Write the expression for the average field due to point charge q at a point which is distance r.

${\mathit{E}}_{\mathbf{a}\mathbf{v}\mathbf{e}}\mathbf{=}\frac{\mathbf{1}}{{\displaystyle \frac{\mathbf{4}}{\mathbf{3}}}\mathbf{\pi }{\mathbf{R}}^{\mathbf{3}}}\mathbf{\int }\mathit{E}\mathit{d}\mathit{\zeta }$ …… (1)

Here, R is the radius of the sphere.

Substitute $\frac{1}{4\pi {\epsilon }_0}\frac{q\hat{}}{r^2}r$for E in equation (1)

$$\begin{gathered} E_{ave}=\frac{1}{\left({\displaystyle \frac{4}{3}}\pi R^3\right)}\int \frac{1}{4\pi {\epsilon }_0}\frac{q\hat{}}{r^2}rd\zeta \\ =\frac{1}{\left({\displaystyle \frac{4}{3}}\pi R^3\right)}\frac{1}{4\pi {\epsilon }_0}\int \frac{q\hat{}}{r^2}rd\zeta \\ =\frac{q}{\left({\displaystyle \frac{4}{3}}\pi R^3\right)}\frac{1}{4\pi {\epsilon }_0}\int \frac{1_{\hat{}}}{r^2}rd\zeta \\ =\rho \frac{1}{4\pi {\epsilon }_0}\int \frac{1_{\hat{}}}{r^2}rd\zeta \left[\boxed{\xcancel{}}\frac{q}{\left({\displaystyle \frac{4}{3}}\pi R^3\right)}=\rho \right] \end{gathered}$$

Solve as further,

$$\begin{gathered} E_{ave}=\frac{1}{4\pi {\epsilon }_0}\int \rho \frac{1_{\hat{}}}{r^2}rd\zeta \\ =\frac{1}{4\pi {\epsilon }_0}\int \frac{\rho \hat{}}{r^2}rd\zeta \\ =E_{\rho } \end{gathered}$$

Hence, $E_{ave}=E_P$is proved.

Determine (b)

b)

Write the expression for the electric filed inside a uniformly charged sphere of the charge density $\rho$.

$E_{\rho }=\frac{1}{3{\epsilon }_0}\hat{\rho }r$ ……. (2)

Substitute $\frac{q}{\left({\displaystyle \frac{4}{3}}\pi R^3\right)}$for $\rho$in equation (2).

$$\begin{gathered} E_{\rho }=\frac{1}{3{\epsilon }_0\left({\displaystyle \frac{-q}{\left({\displaystyle \frac{4}{3}}\pi R^3\right)}}\right)r} \\ =-\frac{q}{4\pi {\epsilon }_0{R}^3}r \\ =-\frac{\rho }{4\pi {\epsilon }_0{R}^3} \end{gathered}$$

Thus, the expression for the electric filed due to dipole moment $R_{\rho }=-\frac{\rho }{4\pi {\epsilon }_0{R}^3}$.

Determine (c)

c)

Let’s consider that, many charges are present inside the sphere.

Then,

Write the expression for sum of the individual average electric field.

$E_{ave}=\frac{-P}{4\pi {\epsilon }_0{R}^3}$

Therefore, the individual average electric field is $E_{ave}=\frac{-P}{4\pi {\epsilon }_0{R}^3}$.

Determine (d)

d)

The charge q is placed outside the sphere at r.

Write the expression for the electric field due to charge.

$E_{ave}=\frac{1}{4\pi {\epsilon }_0}\left(\frac{{\displaystyle \frac{4}{3}}\pi R^3\rho }{r^2}\right)r$

Write the expression for field at P due to uniformly charged sphere.

$R_{\rho }=\frac{1}{4\pi {\epsilon }_0}\frac{-q}{r^2}r$

Hence, average electric filed outside the sphere, due to point charge is equal to electric field is same charge produced at the center of the sphere.